# Prove irrational

1. Mar 13, 2010

### Mentallic

1. The problem statement, all variables and given/known data
I need to prove that $$\sqrt{3}$$ is irrational.

3. The attempt at a solution
The prior problem was to show $$\sqrt{2}$$ is irrational and the solution had to do with a contradiction that each number must be even or something (frankly, I didn't understand it too well). But I don't see how I can apply the same idea to this one, since evenness is not applicable here.

All i can really do so far is assume $$\sqrt{3}=\frac{p}{q}$$ where p and q are integers.
Now, by squaring, $$3q^2=p^2$$.

I am completely lost here though...

2. Mar 13, 2010

### CompuChip

The proof of case with 2 depends on the fact that either p is divisible by 2 (and then you can show that q is too, which contradicts the assumption that you choose them to be "minimal"), or it is not (and since $2q^2 = p^2$ tells you that it is, you have a contradiction).

The proof for 3 is similar: p is either divisible by 3, and then so is q contradicting our assumption; or p is not divisible by 3 and then it must be divisible by 3 which is also impossible.

Perhaps it is a good idea to go through the proof of the irrationality of $\sqrt{2}$ once more, and try to find out what part confuses you. Can you try and explain it to us?

3. Mar 13, 2010

### Mentallic

With the $$\sqrt{2}$$ question the solution says that if $$2q^2=p^2$$ then $$p^2$$ is even, thus p is even, so we can put p=2r for integer r, so now we have $$2q^2=4r^2$$ which also implies that q is even but this is the contradiction.

Actually.. while writing this I tested to see what the difference is for rational cases like $$\sqrt{4}$$ and $$\sqrt{9}$$. Yes, now I understand it. It wasn't the evenness of $$\sqrt{2}$$'s example that was special, it was the fact that it was divisible by 2 (the same thing I know ).

4. Mar 13, 2010

### CompuChip

Yes, in this case it is the same.
But in the general case $\sqrt{n}$, you want to say something about the divisibility by $n$.

(Note that in passing you have also shown that the square of an even number is always divisible by not just 2, but by 4).

5. Mar 13, 2010

### icystrike

If you add an additional information that gcd(p,q)=1 , which also means the greatest common divisor of p and q is 1 . However for $$\sqrt{2}$$ you can draw the conclusion that the gcd(p,q) is atleast 2 and for $$\sqrt{3}$$ gcd(p,q) is atleast 3. In fact you can prove a generalised one analogously which states the $$\sqrt{prime}$$ will lead you to gcd(p,q) is atleast the prime number , yet definition of prime states it cannot be 1 , will leads to the irrationality of all prime.